Question

求不定积分∫1/1+根号(1-x^2)dx

Answer 1

∫dx/(1+√(1-x^2))
x=sinu dx=cosudu √(1-x^2)=cosu
tan(u/2)=sinu/(1+cosu)=x/(1+√(1-x^2))
=∫cosudu/(1+cosu)
=∫[1-1/(1+cosu)]du
=u-∫du/(1+cosu)
=u-∫d(u/2)/(cos(u/2))^2
=u-tan(u/2)+C
=arcsinx - x/(1+√(1-x^2)) +C

追问

tan(u/2)=sinu/(1+cosu)=x/(1+√(1-x^2))请问这一步怎么化的？

追答

tan(u/2)=sin(u/2)/cos(u/2)=[2sin(u/2)cos(u/2)]/[2(cosu/2)^2]=sinu/(1+cosu)
x=sinu cosu=√(1-x^2)
tan(u/2)=x/(1+√(1-x^2))

Answer 2

∫9(cosx)^3)dx=9∫(cos 2;x*cosx)dx =9∫(（1-sin 2;x）*cosx）dx =9∫(cosx-（sin 2;x*cosx)）dx =9∫cosxdx-∫（sin 2;x*cosx)